Zeeman effect eq 1.38 in Foot Atomic Physics

Question

I don't understand how to derive equation 1.38 in Foot's Atomic Physics (preview available on Google Books page 14 here).

I have included the relevant equations below: First we have a differential equation we want to solve:

$$\ddot{\vec{r}} + 2 \Omega_L \dot{\vec{r}} \times \hat{e}_z + \omega_0^2 \vec{r} = 0,$$

where the Larmor frequency is

$$\Omega_L = \frac{eB}{2 m_e}.$$

We look for a solution of the type:

$$\vec{r} = \text{Re} \left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} \exp{(- i \omega t)} \right\}.$$

So far so good.

However when I substitute this into the differential equation, I get:

$$\begin{pmatrix} \alpha \omega^2 & - 2 \Omega_L i \omega \beta & 0 \\ 2 \Omega_L i \omega \beta & \omega^2 \alpha & 0 \\ 0 & 0 & \omega^2 \alpha \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \alpha \omega_0^2 \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

where I define:

$$\alpha = \frac{\exp{(i \omega t) + \exp{(- i \omega t)}}}{2} $$

and

$$\beta = \frac{\exp{(i \omega t) - \exp{(- i \omega t)}}}{2} $$.

This almost looks like eq 1.38 in the book:

$$\begin{pmatrix} \omega_0^2 & - 2 \Omega_L i \omega & 0 \\ 2 \Omega_L i \omega & \omega_0^2 & 0 \\ 0 & 0 & \omega_0^2 \alpha \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \omega^2 \begin{pmatrix} x \\ y \\ z \end{pmatrix},$$

where the two differences are $\omega_0 \to \omega$ on the RHS and setting $\alpha = \beta$.

I cannot see where I am going wrong, and the book does not mention any sort of approximation that might make $\alpha = \beta$. Intuitively if the solution type we are assuming did not include the real part only, I can see how the exponential terms would cancel. But since it includes the real part, then letting $z = a + bi$ be an arbitrary complex number: $$\text{Re}(z) = a = \frac{a + bi + a - bi}{2} = \frac{z + \overline{z}}{2},$$ we should get two exponentials. Then taking one derivative changes the sign on one of the exponentials.

Answer 1

When you use complex wave notation, with the ansatz $$ \vec{r} = \Re \left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} \exp{(- i \omega t)} \right\}. $$ you don't actually substitute in $\Re(x e^{-i \omega t}) = \frac{x}{2} (e^{i \omega t} + e^{-i \omega t})$. Instead, you just substitute in $x e^{-i \omega t}$. See this answer for an exposition of what we're "really" doing when we use this notation and why it works.

If you substitute this in with just the $e^{i \omega t}$ term, you get $$ \begin{pmatrix} \omega_0^2 & - 2 \Omega_L i \omega & 0 \\ 2 \Omega_L i \omega & \omega_0^2 & 0 \\ 0 & 0 & \omega_0^2 \end{pmatrix} \begin{pmatrix} x e^{-i \omega t} \\ y e^{-i \omega t} \\ ze^{-i \omega t} \end{pmatrix} = \omega^2 \begin{pmatrix} xe^{-i \omega t} \\ ye^{-i \omega t} \\ ze^{-i \omega t} \end{pmatrix}, $$ and in this form it is obvious that the complex exponentials cancel out, leaving the form from the textbook you're working from.